# Example of a non-uniformly continuous function (TIFR 2013 problem 19)

Question:

True/False?

Every differentiable function $$f:(0,1)\to [0,1]$$ is uniformly continuous.

Hint:

$$sin(1/x)$$ is not uniformly continuous. However, range is not [0,1]. But its a simple matter of scaling.

Discussion:

Let $$f(x)=sin(\frac{1}{x})$$ for all $$x\in(0,1). For simplicity, we first prove that this f is not uniformly continuous. Then we will scale things down, which won’t change the non-uniform continuity of f. Note that f is differentiable. To show that f is not uniformly continuous, we first note that as \(x$$ approaches 0, $$1/x$$ goes through an odd multiple of $$\frac{\pi}{2}$$ to an even multiple of  (\frac{\pi}{2}\) real fast. So in a very small interval close to 0, I can find two such points which gives value 1 and 0.

Let $$x=\frac{2}{(2n+1)\pi}$$ and $$y=\frac{2}{2n\pi}$$.

Then $$|x-y|=\frac{2}{2n(2n+1)\pi}$$.

Since the right hand side of above goes to zero as n increases, given any $$\delta > 0$$ we can find n large enough so that $$|x-y|<\delta$$. For these x and y, $$|f(x)-f(y)|=1$$.

Of course, choosing $$\epsilon$$ as any positive number less than 1 shows that f is not uniformly continuous.

We have with us a function $$f$$ which is bounded, differentiable and not uniformly continuous.

To match with the questions requirements, notice $$-1\le f(x)\le 1$$.

So $$0\le 1+f(x) \le 2$$. And $$0\le \frac{1+f(x)}{2} \le 1$$.

Define $$g(x)= \frac{1+f(x)}{2}$$ for all $$x\in(0,1)$$.

Since sum of two uniformly continuous functions is uniformly continuous and a scalar multiple of uniformly continuous function is uniformly continuous, if $$g(x)$$ was uniformly continuous, then $$f(x)=2g(x)-1$$ would also be uniformly continuous.

Which proves that g is in fact not uniformly continuous. It is still differentiable, and range is $$[0,1]$$, which shows that the given statement is actually false.